Question #10e47

2 Answers
Jan 18, 2018

Toque applied = 6N.m & power is about120 watt

Explanation:

Here,by applying the torque angular retardation was done to stop its motion(considering no skidding was there).

so, angular retardation= α = change in angular velocity i.e ωt = 4010 i.e 4 radian/s2

Now, torque =Iα (where, I is the moment of inertia)

moment of inertia of a cylinder passing through its axis is Mr22 =1.5Kg.m2

so. τ =1.54N.m i,e 6N.m

so,angular power required = τθt i.e about 120 watt(τθ= work done,dividing by time required for this angular displacement of θ by the time t in which it was brought to rest due to application of this torque) ( θ is angular displacement)(θ=(ω)22α=40224=200radian)

Jan 18, 2018

The torque is =6Nm. The power is =240W

Explanation:

The torque and the angular acceleration are related

Torque (Nm)=Moment of inertia (kgm^2)×angular acceleration (rads^-2)

τ=I×α

The moment of inertia of the solid cylinder is

I=mr22

The mass of the cylinder is m=3kg

The radius of the cylinder is r=1.0m

The moment of inertia is I=3×122=1.5kgm2

The initial angular velocity is ω0=40rads1

The final angular velocity is ω1=0rads1

The time is t=10s

Apply the equation (to find the angular acceleration (α))

ω1=ω0+αt

α=ω1ω0t=04010=4rads2

The torque to be applied is

τ=Iα=1.5×4=6Nm

Power (W)=torque (Nm)×angular velocity (rads^-1)

P=τ×ω0

The power is P=6×40=240W